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E, B and the direction of propagation of the electromagnetic wave is always perpendicular to one another

The speed of electromagnetic waves

Electromagnetic waves such as light are transverse waves. However, it is not particles which oscillate, but an electric field (E) coupled with a magnetic field B , as shown above. The speed of the waves through space, called the speed of light,c depends on the permittivity,e0 and permeability m0, of a vacuum :

speed of light (in vacuum) = 3.0 x 108 m s-1

25.2 Relationship between e0, m0 and c

The relationship of e0 ,m0 and c

Velocity of electromagnetic wave, c =

where

e0 = 8.85 x 10-12 F m-1 (Free space permittivity) and

m0 = 4p x 10-7 H m-1 (Free space permeability)

c = 2.998 x108ms-1

Example 1

The velocity of an electromagnetic waves in vacuum is given by the equation

(a) Calculate this velocity.

(b) What deduction can be made from this information?

(c) What is it that oscillates in the electromagnetic waves?

25.3 Electromagnetic wave spectrum

The electromagnetic spectrum

Light Is one member of a whole family of transverse waves called the electromagnetic spectrum. In empty space, these waves all travel at the same speed: 300 000 km s-1 . Electromagnetic waves are emitted whenever electrons or other charged particles oscillate or lose energy. The greater the energy change, the lower the wavelength.

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Electronic circuits handle small, changing electric currents. The changes are called signals. The diagram below shows the main parts of a basic electronic system.

Note:

•Devices which change signals from one form to another (e.g. sound to electrical) are called transducers(a device that converts one type of energy to another).

•Input signals can be very weak. But by means of electronic circuits, they can control output signals which can be much stronger. The output power is provided by the supply (usually low voltage DC).

Two possible functions of electronic circuits are described below.

1. Switching Depending on the input signals received, the output voltage is either HIGH (close to the supply voltage) – or LOW (zero), so the output device is either ON or OFF.

2. Amplification The output signals are an amplified (magnified) version of the input signals. For example, very low voltage AC from a microphone causes a higher voltage AC output for a loudspeaker.

The voltagegain of an amplifier is defined like this:

Note:

•In electronics, the term ‘voltage’ is commonly used for potential, PD, and EMF.

The key component in electronics is the transistor. It can amplify or act as a switch. An integrated circuit (IC), in a package as on the right, may have thousands of transistors and other components formed on a single chip of silicon.

Input devices

Input transducers (e.g. microphones) are called sensors. They must be linked to the circuit in such a way that any change they detect (e.g. a pulse of sound) causes a change in input voltage.

Sensors generating a voltage These include some microphones, and those light sensors that work like solar cells . Ideally, the voltage is in proportion to the external change causing it.

Sensors with varying resistance These include:

• light-dependent resistors (LDRs) ,

•thermistors (temperature-dependent),

•strain gauges (resistance changed by stretching).

To produce the necessary voltage change, the sensor forms one part of a potential divider . For example, when light falls on the LDR below, its resistance falls. The LDR therefore takes a smaller share of the supply voltage, so the voltage V falls.

Output devices

These include loudspeakers, buzzers, LEDs, and relays. Light-emitting diodes (LEDs) These emit light when a small current passes through. They can be used as indicator lamps to show the presence of an output voltage. Like all diodes, they only conduct in one direction (see D1 and D2).

To avoid damage, a resistor must be placed in series with an LED to limit the current through it.

The LED on the right can take a current of 0.01 A, which produces a 2 V drop across it. If a circuit’s output voltage is 9 V, there needs to be a 7 V voltage drop across the resistor, so its resistance should be 7/0.01 = 700 S2.

Relays These are electromagnetic switches. A small current activates an electromagnet. This closes (or opens) a switch which can control the flow of a larger current in a separate circuit, e.g. a circuit with a mains heater or motor in it.

Impedances and matching

Resistance is one form of impedance (see D4). The microphone and loudspeaker below each have impedance. The amp[ifer has input impedance and output impedance rather as a battery has internal resistance .

In the diagram, the various impedances are represented by resistors in the input and output circuits.

In each circuit, it is the higher impedance that has the higher voltage drop across it. So

• For maximum input voltage, the microphone’s impedance should be as low as possible.

• For maximum output voltage, the loudspeaker’s impedance should be as high as possible. However, this gives a very low output current and almost no sound!

• For maximum output power, the loudspeaker’s impedance should be the same as the output impedance. (Output voltage x output current is then at its highest.)

A transformer can be used to match impedances and give maximum power transfer. A transformer with an impedance Z in its primary circuit is equivalent to an impedance of Z(N2/N1)2.

The circuit diagram shows an op-amp with an open loop gain of 105 connected to a ±9V voltage supply. Sketch graph to show the variation of the input voltage Vin and output voltage Vout with time t for each of the following input voltages.

(a) sinusoidal a.c voltage of peak value 20mV

(b) sinusoidal a.c voltage of peak value 120 mV

(c) sinusoidal a.c voltage of peak value 2.0 V

24.2 Op amp (close loop)

Differential amplifier – it amplifies the voltage between inverting and non inverting, so V output given by

Open loop gain (gandaan voltan gelung-buka),

In the case of op amp, A0 is depend on the frequency.

To reduce gain(actually to reduce frequency), op amp are used with a resistor or wire linking the output to one input.

When connecting output to the inverting input. The connection is called with another name as

Negatif feed close loop back in short it just called a Negatif feedback.

Negative feedback

The amplifier circuit on the right uses a closed loop to feed back a set fraction of the output voltage to the inverting input. This is called negative feedback because the signal being fed back partly cancels the input signal.

Advantages using negatif feeback

voltan output tidak sefasa dengan voltan input

mengurangkan voltan input setelah ianya disuap balik

reducing gain (voltage gain)

Advantages of reducing gain (Using closed loop op amp) These include the following:

• The bandwidth is greater, i.e. the gain is constant over a wider range of input frequences (see below).

Note : Bandwidth = frequency which the gain is constant

Graf A0 vs frekuensi

24.3 Inverting and non-inverting amplifiers

V0 = A0 (V2 – V1) where A0= open loop gain

1 Inverting amplifier

Gain

Explanation:

Op amp as an inverting amplifier

The amplifier circuit on the right uses a closed loop to feed back a set fraction of the output voltage to the inverting input. This is called negative feedback because the signal being fed back partly cancels the input signal.

To calculate the amplifier’s closed-loop voltage gain the following assumptions are made:

• The op amp is not saturated. It has such a high open-loop gain that the voltage difference between its inverting and non-inverting inputs must be negligible. Therefore P is effectively at 0 V. (P is called a virtual earth.)

• The current through P is negligible. Therefore, the current through Rf is the same as through Ri (see D3).

If I is the current through Rf (feedback resistor) and Ri,

voltage drop across Rf = IRf

voltage drop across Ri = IRi

But if P is at 0V,

voltage drop across Rf = 0 – Vout = -Vout

voltage drop across Ri = Vin – 0 = Vin

From the above, it follows that

So closed-loop voltage gain =

From the closed-loop voltage gain equation below:

• An inverting amplifier has a negative gain.

• The gain does not depend on the characteristics of the op amp. It is set by the values of Rf and Ri. For example, if Rf = 200 kW and Ri = 20 kW, the gain = -200/20 = -10.

Contoh 2

The circuit diagram shows an inverting amplifier the power supply is ± 9V.

(a) what is the voltage gain of the circuit ?

(b) what is the output voltage V0.

(i) x = 200 kW and Y = 50 kW

(ii) x = 50 kW and Y = 150 kW

2 Non-inverting amplifier

Gain

Explanation:

Op amp as a non-inverting amplifier

In the circuit on the top, the input signal goes to the non-­inverting input, and the potential divider provides negative feedback. With the same assumptions as above,

closed-loop voltage gain

A non-inverting amplifier has a much higher input impedance than the inverting type. This is useful, for example, where a high-impedance microphone is connected to the input (see Impedances and matching in H16).

Voltage follower This is a non-inverting amplifier in which Rf is zero (i.e. a direct connection) and Ri is infinite (i.e. removed). In this case, the gain is 1, so the output voltage is the same as the input voltage. However, the output impedance is much less than the input impedance, which makes the circuit useful for impedance matching.

Contoh 3

In the circuit diagram shown, the input signal is Vin and the output signal is V0

(a) state and explain the type of amplifier

(b) what is the output voltage V0 if Vin

(i) +1.5 V

(ii) +2.5 V

(c) sketch a graph to show the variation of the output voltage V0 with time t, if

24.4 Use of operational amplifiers

3 Uses of op-amps

(a) Changing sinusoidal wave forms to rectangular waveforms

(b) Automatic electrical switch/ Comparator

On the right, the LED is switched on automatically when darkness falls. The switching is done by an op amp which compares the voltages from two potential dividers:

V– is fixed by the ratio of R2 to R1,

V+ varies with the resistance of the LDR.

There is no feedback, so a small difference between V– and V+ saturates the op amp, causing maximum output voltage.

In bright light, the LDR has a low resistance, so V+ is low and less than V– . The op amp is saturated, but its output voltage is negative, so the LED cannot conduct.

As the light intensity falls, the resistance of the LDR rises, and so does V+. When V+ > V– the op amp saturates in the opposite direction. The output voltage is now positive, so the LED conducts and lights up.

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Above, a magnet is moved into a coil. If the flux through the coil changes (at a steady rate) by DF in time Dt, then an EMF of is induced in each turn. But there are N turns in series. So, the total induced EMF E is as follows:

For example, if the flux changes by 6 Wb in 2 s, and the coil has 100 turns, then the total induced EMF is 300 V.

Above, a current-carrying wire is at right-angles to a uniform magnetic field. The field exerts a force on the wire. The direction of the force is given by Fleming’s left-hand rule . The size of the force depends on the current I, the length l in the field, and the strength of the field. This effect can be used to define the magnetic field strength, known as the magnetic flux density, B:

FB = BIl (1)

B is a vector. The SI unit of B is the testa (T). For example, if the magnetic flux density is 2 T, then the force on 2 m of wire carrying a current of 3 A is 2 x 2 x 3 = 12 N.

If a wire is not at right angles to the field, then the above equation becomes

FB=BIl sinq

where q is the angle between the field and the wire. As q becomes less, the force becomes less. When the wire is parallel to the field, sin q = 0, so the force is zero.

B inside a solenoid with a core The value of B is changed by a core. For example, with a pure iron core, B is increased by a factor of about 1000 (depending on the temperature). The previous equation then becomes:

B = µrµ0 nI

µr, is called the relative permeability of the material. So, for pure iron, µr, is about 1000.

An electromagnet is a solenoid with a core of high µr.

21.5 Force between two current-carrying conductors

Force between two current-carrying wires

X and Y above are two infinitely long, straight wires in a vacuum. The current in X produces a magnetic field, whose flux density is B at Y. As a result, there is a force on Y. F is the force acting on length l.

From equation (3)

From equation (1) FB = BI2 l

Note:

• The above equation gives the force of X on Y. Working out the force of Y on X gives exactly the same result.

• If the two currents are in the same direction (as above), then the wires attract each other. If the two currents are in opposite directions, then the wires repel.

One ampere is the current which, flowing through two infinitely long, thin, straight wires placed one metre apart in a vacuum, produces a force of 2 x 10-7 newtons on each metre length of wire.

Using the various factors in equation

,

the above definition can be expressed in the following form (for simplicity, units have been omitted):

If I1=I2=1A,a=1m and l=1m,then F= 2×10-7N. Substituting these in equation above gives µ0 = 4p x 10-7. The above definition is not a practical way of fixing a standard ampere. This is done by measuring the force between two current-carrying coils.

Current Balance

– Measure current by using the principle of the force between two paralell wire using mechanical

Prinsip:

m = jisim pemberat W

b = jarak F ke tuas

c = jarak berat ke tuas

L = jarak XY

apabila sistem tuas seimbang kita dapati

Known that

FB = BIL

B=µ0nI

Maka FB = µ0n I . IL

Then

21.7 Torque on a coil

Lets consider the force on PQ and RS that involve in producing the torque.

Force that produce on PQ and RS have the same magnitude but different direction.

so force,F

where

L = height of coil(length of the coil in the magnetic field) and N = no. of turns carrying current,I.

Torque

where

q = angle (below the horizontal from the plan view) between magnetic flux density and the coil.